I'm hoping to really knock out several questions I have in my mind with just this one. I've been doing a lot of practice problems on this topic, and although I get the right answers, I really don't know what the answers mean. So there's a theorem that says

If F is a vector field defined on all
of $R^3$ whose component functions
have continuous partial derivatives and
curl F=0, then F is a conservative
vector field.

So that's great, but it really doesn't give me an understanding of what that really means.

First of all, what does it mean for all of its component functions to have continuous partial derivatives. I mean I know how to determine if they are or not, but what does it mean if they are all continuous or if some are not, and how does that change what a conservative field is?

Second, what does it mean that the curl F=0, and why is it such an important occurrence that we gave it a special name such as conservative?

And lastly, (this I should know but sadly not), is there a difference in a vector field being defined on all of $R^3$ compared to a vector field being continuous on all of $R^3$. Do those mean the same thing?

$\begingroup$the "defined on all of $\mathbb{R}^3$" matters because topological considerations become important (ie the statement might not be true for various subsets of $\mathbb{R}^3$ say with "holes" in them).$\endgroup$
– yoyoMar 31 '11 at 20:34

$\begingroup$What is the definition of "conservative" that you are using? There are two or three (equivalent) ones in common use.$\endgroup$
– Jesse MadnickMar 31 '11 at 21:16

$\begingroup$@JesseMadnick if they are equivalent, why is it relevant?$\endgroup$
– pppqqqFeb 11 '14 at 17:57

$\begingroup$@pppqqq: Because knowing where the OP is coming from (in terms of a knowledge base) can help us craft answers better suited to the OP. Also, some definitions of "conservative" might be more intuitive or geometric than others.$\endgroup$
– Jesse MadnickFeb 11 '14 at 18:05

2 Answers
2

Let $f$ be a function on some set $A$. When we say that $f$ is defined on $A$, we mean that $f(x)$ exists for every $x$ in $A$. However, just because $f$ is defined on $A$ doesn't mean that it is continuous on $A$: it may be continuous, or it may not be. The textbook intuition of continuity is that $f$ is continuous on $A$ if we can draw its graph without picking up our pencil.

Note, though, that to even talk about continuity on a set, our function has to be defined there first.

Now vector fields are really just a special type of function (one that inputs points and outputs vectors), so all of this applies to vector fields, too.

Before I address Questions 1 and 2, let's first distinguish between a few concepts.

Let $F$ be a vector field defined, continuous, and having continuous partial derivatives on a region $U$ in $\mathbb{R}^3$. Consider the following four properties that $F$ may have:

(1) (Path-independence.) The line integral of $F$ between two points does not depend on the path chosen. That is, if $C_1$ and $C_2$ are paths between $a$ and $b$, then $$\int_{C_1} F\cdot dr = \int_{C_2} F\cdot dr.$$

(2) (Integrals of closed loops are zero.) The line integral of $F$ around any closed curve is zero. That is, if $C$ is a closed curve, then $\int_C F\cdot dr = 0$.

Fact 3: If in addition $F$ is defined on all of $\mathbb{R}^3$ and has continuous partial derivatives, then property (4) implies the first three.

In other words, if $F$ is defined on all of $\mathbb{R}^3$ (and has continuous partial derivatives), then all four concepts coincide.

However: As yoyo mentioned in the comments, property (4) is not in general equivalent to conservativity. The classic example is the vector field on $\mathbb{R}^2 \setminus (0,0)$ given by
$$F(x,y) = \left(\frac{y}{x^2 + y^2}, \frac{-x}{x^2 + y^2} \right).$$
This vector field has the curious property that $\text{curl }F = 0$, yet does not satisfy any of the three (equivalent) conservativity properties. The reason for this is that $F$ is not defined on all of $\mathbb{R}^2$ because it is undefined at the origin $(0,0)$.

There are similar examples for $\mathbb{R}^3 \setminus (0,0)$ but I can't seem to come up with any right now.

So now we can address Question 2. As I mentioned above, conservativity is not the same as saying that $\text{curl F} = 0$. However, if $F$ is defined on all of $\mathbb{R}^3$, then yes, they coincide.

So why is $\text{curl }F = 0$ such a good property to have? Well, the easy answer is that: (a) it's an easy property to check, and (b) in the event that $F$ is defined on all of $\mathbb{R}^3$, then we get the three (equivalent) conservativity conditions, which I hope you can see are good things to have.

The physical intuition behind conservativity is that it models conservative forces like gravity. In this scheme, $F$ would represent the force and $\int_C F\cdot dr$ would represent the work (energy) of applying $F$ to some object over a path $C$. It is intuitively clear that the work done by gravity (say) over any closed loop is zero: the potential energy hasn't changed any.

Finally, let me address Question 1. Technically speaking, to talk about conservativity, you're right, we don't actually need the partial derivatives of $F$ to be continuous, or even to exist. That is, we can talk about properties (1), (2), and (3) above without the assumption that the partial derivatives of $F$ exist.

However, to talk about property (4), where we take the curl, we need to say that the partial derivatives of $F$ exist. Do we need to assume in addition that the partial derivatives are continuous? My guess is probably not (someone please correct me if I'm wrong). But assuming that the partial derivatives are continuous is certainly a nice simplifying assumption to have.

$\begingroup$It is indeed true that simple differentiability is sufficient. Continuous derivative is not necessary, although may be easier to check.$\endgroup$
– scineramMar 31 '11 at 22:35

$\begingroup$@scineram: By "differentiability" do you mean Frechet differentiability (i.e. the total derivative existing), or do you mean that all the partial derivatives exist?$\endgroup$
– Jesse MadnickMar 31 '11 at 23:07

You should try to use some physical intuition here. Think of the vector field as being the force acting on some object. A "force field" being conservative means that if you pick two points at random, then the work needed to move that object from one point to the other is completely independent of the path it takes.

You could push it along a huge curvy path, or just take the straight line path. It doesn't matter, the work (integral of the force over the path) will be the same. That's conservative.

Curl is some sort of measurement on how much the vector field spirals around. Taking that more seriously than it probably should be taken, if you have the force field literally go around in a circle (for every point on the circle the vector at that point is tangent, draw this to get the idea), then it does not have 0 curl and moreover we easily see that (using physical intuition) that the amount of work to needed to go around the circle one way will be a lot and the other way will be none, so it isn't conservative.

Of course, that is not equivalent to the statement you put up, but now you can think if it isn't curving around anything, then the above situation can't happen and if that's the only way you get differing work then that is the physical explanation of the theorem.